# What is logistic growth in ecology?

##### 1 Answer
Mar 30, 2015

It's often used as a population model and it's a step up in complexity and accuracy from an exponential growth model. An exponential growth model assumes the relative growth $\setminus \frac{\setminus \frac{\mathrm{dP}}{\mathrm{dt}}}{P}$ of a population is constant, which is unrealistic in the long run, whereas logistic growth assume the relative growth $\setminus \frac{\setminus \frac{\mathrm{dP}}{\mathrm{dt}}}{P}$ can be thought of as a linearly decreasing function of $P$, say $\setminus \frac{\setminus \frac{\mathrm{dP}}{\mathrm{dt}}}{P} = k \cdot \left(1 - \setminus \frac{P}{L}\right)$, so that $\setminus \frac{\mathrm{dP}}{\mathrm{dt}} = k P \left(1 - \setminus \frac{P}{L}\right)$. The constant $L$ is called the carrying capacity of the environment.

If you solve this differential equation using separation of variables and the initial condition $P \left(0\right) = {P}_{0}$, you'll get $P \left(t\right) = \setminus \frac{{P}_{0} L {e}^{k t}}{{P}_{0} {e}^{k t} + L - {P}_{0}} = \setminus \frac{L}{1 + \setminus \frac{L - {P}_{0}}{{P}_{0}} {e}^{- k t}}$. This can also be written as $P \left(t\right) = \setminus \frac{L}{1 + {e}^{- k \left(t - {t}_{0}\right)}}$, where \frac{L-P_{0}}{P_{0}}=e^{kt_{0}. Note that $P \left({t}_{0}\right) = \setminus \frac{L}{2}$.

For $0 < {P}_{0} < L$, the graph of $P \left(t\right)$ is an increasing function of $t$ that looks somewhat like an elongated "S" (the technical name for the shape is "sigmoidal") with $\setminus {\lim}_{t \setminus \rightarrow \setminus \infty} P \left(t\right) = L$. It turns out that ${t}_{0} = \setminus \frac{1}{k} \ln \left(\setminus \frac{L - {P}_{0}}{{P}_{0}}\right)$ is the time when the graph of $P \left(t\right)$ has an inflection point (changing from concave up to concave down). See the figure below.